Question

A baised coin with probability $p(0<p<1)$ of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is $\frac{2}{5}$, find $3p$.

Answer 1

Let $X$ denote the number of tosses required.
Then $P(X=r)=(1-p{)}^{r-1}p$ for $r=1,2,3,...$
Let $E$ denote the event that the number of tosses required is even. Then $P\left(E\right)=P\left[\right(X=2)\cup (X=4)\cup (X=6)\cup ...]$

$=P(X=2)+P(X=4)+P(X=6)+...$

$=(1-p)p+(1-p{)}^{3}p+(1-p{)}^{5}p+....$

$=\frac{p(1-p)}{1-(1-p{)}^2}=\frac{p(1-p)}{2p-p^2}=\frac{1-p}{2-p}$

As we are given that $P\left(E\right)=\frac{2}{5}$, we get

$5(1-p)=2(2-p)\Rightarrow 1-3p=0$

$\Rightarrow p=\frac{1}{3}\Rightarrow 3p=1$